Re: Differentiability of f everywhere but |f| nowhere, f: R - > R?

In article <[EMAIL PROTECTED]>,
painch1207 <[EMAIL PROTECTED]> wrote:

>Exhibit a differentiable function f:R->R such that the function |f|
>defined by

>|f|(t) = |f(t)|

>is not differentiable.  (Spivak, Calculus on Manifolds 2-13d)

>It is a little bit mystifying to me exactly what he is looking for
>here.  Unless the problem is requiring that the derivative exists
>nowhere on R it seems uniquely unchallenging (i.e. very unlike a
>Spivak problem, for the most part).

>So I was wondering if anyone knew of such a function that is as
>described above, with the assumption that |f|'(t) exists nowhere on R.

No: it's obvious for f:R->R that |f| is differentiable at any point 
where f is differentiable and nonzero.  On the other hand, if f is 
everywhere 0 then |f| is again differentiable.  I suspect that the
point of having this after (a-c) is that the statement "|f| is 
differentiable when f is differentiable and nonzero" is also true for
f:R -> R^n.

Robert Israel                                [EMAIL PROTECTED]
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2